There is no hard and fast rule. Consider that in some LC circuit applications, a lower Q may be desirable in order to achieve a wider bandwidth. In other cases, a very high Q may be desirable for narrow selectivity, for example.
Both the inductor and the capacitor in a resonant circuit may affect the Q of the circuit.
The Q of the inductor is determined by its inductive reactance divided by its series resistance.
$$Q_L=\frac{X_L}{R_L} \tag1$$
Since inductance is generally a factor of turns squared and since the resistance of the inductor is a factor of turns, this first order analysis indicates that the lower the inductor value for a given wire type/diameter and for a given construction method, the higher the Q of the inductor. The higher the Q of the inductor, the higher the Q of the resonant circuit.
The Q of a capacitor is determined by its capacitive reactance divided by its effective series resistance.
$$Q_C=\frac{X_C}{ESR_C} \tag2$$
In most practical cases, the ESR of the capacitor is a factor only in series resonant circuits. In a parallel resonant circuit, generally the series resistance of the inductor will dominate the Q.
In a series resonant circuit, the resistive losses of the inductor and capacitor are simply added. Since you quoted the formula for a series resonant circuit, this should be your approach.
$$Q=\frac{1}{R_L+ESR_C} \sqrt{\frac{L}{C}} \tag3$$
Before selecting a final inductor value, make certain that its self resonance will not negatively affect your circuit.
The insertion loss under this scenario is given as:
$$ \text{Insertion Loss} = 20\log\left({1-\frac{Q}{Q_L}}\right) \tag 4$$
where QL is as noted above and Q is the series circuit Q as noted in the equation in your question.
Other factors that may come into play, depending upon the application, are the inductance and the Q of the inductor over a wide frequency range; the stability of the capacitor over a wide temperature range; and the size, weight, tolerance, cost and availability of the components.